Group 4

TASK 1

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Assuming that the 2 men are on earth, the building is perpendicular to the ground, the 2 men and the building are on the same flat surface, the angle of elevation is measured from their feet, the building is viewed at the same time and that it is a 2D problem (if in 3D, there would be infinite posibilities)

Let x be the distance between the first man and the building (30˚ elevation) and y be the distance between the second man and the building (60˚ elevation)

tan30˚ = 120/x

x = 120/tan30˚ = 207.8m (4sf)

tan60˚ = 120/y

y = 120/tan60˚ = 69.28m (4sf)

Case 1: The 2 men are standing on the opposite sides of the building.

Therefore,

Distance between the 2 men = x + y = 277m (3sf)

Case 2: The 2 men are standing on the same side of the building.

Therefore,

Distance between the 2 men = x - y = 139m (3sf)

TASK 2A

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Sine Rule

Sine rule sinA/a = SinB/b = sinC/c

Based on the diagram:

side a is opposite angle A, side b is opposite angle B

sin A = h/b
sin B = h/a

Make h the subject

bsin A = h
asin B = h

Therefore: bsin A = asinB

Divide both sides by ab

sinA/a = sinB/b

This rule can be used to find unknown lengths or angles in a triangle as long as we have at least two sides one angle (to find unknown angle) or two angles one side (to find unknown side) t

Ambiguous Cases

Consider this example:

A triangle PQR where side PQ is 41cm, side PR is 28cm and angle PQR is 39˚. Find the angle at R.

At first glance, the triangle can look like this:

If we try to find the angle at R, we get

sin39˚/28 = sinR/41

sinR = 41 * (sin39˚/28) = 0.923 (3sf)

Therefore angle R = 67.1˚

HOWEVER

sin112.9˚ is also 0.923 (3sf)

Using ASTC, we find that sin 67.1˚ is equal to sin112.9˚

Therefore, angle R can also be 112.9˚ and using the given information, the triangle could also look like this:

Both triangles are valid according to the information given in the question. But since no diagram was given, there are 2 possible answers the this question. Therefore, this is an ambiguous case.

TASK 2B

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Cosine Rule

It is a generalization of the Pythagoras formula .
Pythagoras formula states that a^2+b^2=c^2 . The cosine rule is a^2=b^2+c^2-2bcCosA
The reason why I say that it is a generalization of the Pythagoras theorem is because when we try to use it with a right angled triangle ,

If we put in the values of the equation , we get this
c^2=a^2+b^2-2abCos90
As Cosine 90 is 0 ,
c^2=a^2+b^2 . 
Therefore , the cosine rule is related to Pythagoras theorem .

1. Using cosine rules for right angled triangles , cos A=x/b and cosB=h/(c-x)
2. Use manipulation to change the equation into x=bcosA
3.We can apply Pythagoras theorem to the 2 triangles in the diagram , (c- x)^2+h^2=a^2 for the left triangle and x^2+h^2=b^2
4. As they both have the common term of h^2 , we can substitute them with each other .

(c-x)^2-a^2=x^2-b^2 

5. Simplify the equation:

c^2-2cx+x^2-a^2=x^2-b^2

c^2-2cx-a^2=-b^2 

b^2=a^2-c^2+2cx

6. We can also substitute the equation from the second step into the equation . b^2=a^2-c^2+2bcCosA
7. There are multiple ways to rearrange this equation and the most common way is a^2=b^2+c^2-2bcCosA 

TASK 3

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TRIGONOMETRY - BEARING

Let the fire be point C

Assuming the the 2 towers are looking at the same point in the fire.

Angle C = 180˚ - 70˚ - 40˚ = 70˚

Using sine rule:

sinC/c = sinB/b = sinA/a

sin70˚/7 = sin70˚/b = sin40˚/a

b = sin70˚/(sin70˚/7) = 7km

a = sin40˚/(sin70˚/7) = 4.79km

Therefore, tower B is closer to the diagram by 7km - 4.79km =  2.21km